Consider the following two subspaces of \(\mathbb{R}^4\) \[ W_1=\{(x_1,x_2,x_3,x_4):x_1+x_2+x_3+x_4=0\} \] \[ W_2=\{(x_1,x_2,x_3,x_4):x_1+2x_2+3x_3+4x_4=0\}. \] Which of the following statements is correct?
Show Hint
Two independent linear equations in 4 unknowns cut down the solution space by 2 dimensions, so the intersection of the two hyperplanes has dimension 4-2=2.
Step 1: Think of $W_1$ and $W_2$ as hyperplanes with normal vectors.
$W_1$ is the set of vectors orthogonal to $n_1=(1,1,1,1)$, and $W_2$ is the set of vectors orthogonal to $n_2=(1,2,3,4)$. Each hyperplane in $\mathbb{R}^4$ has dimension $3$, since one equation removes one degree of freedom.
Step 2: Use the dimension formula for an intersection of two hyperplanes.
For two hyperplanes through the origin in $\mathbb{R}^n$ given by independent normal vectors, \[ \dim(W_1\cap W_2)=n-2 \] as long as $n_1$ and $n_2$ point in genuinely different directions (are linearly independent). Here $n=4$.
Step 3: Check that $n_1$ and $n_2$ are independent.
$n_1=(1,1,1,1)$ and $n_2=(1,2,3,4)$ would be dependent only if $n_2=k\,n_1$ for some scalar $k$. Matching the first entries gives $k=1$, but then $n_1$ scaled by $1$ would need every entry equal to $1$, while $n_2$ has entries $1,2,3,4$ that are not all equal. So no such $k$ works, and $n_1,n_2$ are linearly independent.
Step 4: Apply the formula.
\[ \dim(W_1\cap W_2)=4-2=2 \]
Step 5: Confirm with an explicit parametrization.
Solving $x_1+x_2+x_3+x_4=0$ and $x_1+2x_2+3x_3+4x_4=0$ together, subtract to get $x_2+2x_3+3x_4=0$, so $x_2=-2x_3-3x_4$, and then $x_1=x_3+2x_4$. Treating $x_3,x_4$ as free parameters gives a 2-dimensional family of solutions, confirming the count.
Step 6: Compare with the answer choices.
A dimension of $0$ or $1$ would need the normals to force more independent constraints than we actually have, and a dimension of $3$ would need $n_1$ and $n_2$ to be scalar multiples of each other, which Step 3 ruled out. Only dimension $2$ fits.
\[ \boxed{\dim(W_1\cap W_2)=2} \]
